Problem: Simplify the following expression and state the condition under which the simplification is valid. $t = \dfrac{y^3 + 11y^2 + 30y}{7y^3 + 42y^2 + 35y}$
First factor out the greatest common factors in the numerator and in the denominator. $ t = \dfrac {y(y^2 + 11y + 30)} {7y(y^2 + 6y + 5)} $ $ t = \dfrac{y}{7y} \cdot \dfrac{y^2 + 11y + 30}{y^2 + 6y + 5} $ Simplify: $ t = \dfrac{1}{7} \cdot \dfrac{y^2 + 11y + 30}{y^2 + 6y + 5}$ Since we are dividing by $y$ , we must remember that $y \neq 0$ Next factor the numerator and denominator. $ t = \dfrac{1}{7} \cdot \dfrac{(y + 5)(y + 6)}{(y + 5)(y + 1)}$ Assuming $y \neq -5$ , we can cancel the $y + 5$ $ t = \dfrac{1}{7} \cdot \dfrac{y + 6}{y + 1}$ Therefore: $ t = \dfrac{ y + 6 }{ 7(y + 1)}$, $y \neq -5$, $y \neq 0$